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Chapter 2: Problem 8

Let \(A\) be a nonsingular \(n \times n\) matrix with \(n>1\) Show that \\[ \operatorname{det}(\operatorname{adj} A)=(\operatorname{det}(A))^{n-1} \\]

Short Answer

Expert verified
The determinant of the adjugate of a nonsingular n x n matrix A, denoted as det(adj(A)), is equal to the determinant of A raised to the power of n-1: \[ \det(\text{adj}(A)) = (\det(A))^{n-1} \]

Step by step solution

01

Definitions

Recall that the adjugate of a matrix A is the transpose of the cofactor matrix of A, denoted as adj(A). The cofactor matrix of A, denoted as C, is a matrix with the same size as A and its elements are the cofactors c_ij of A. The cofactor c_ij is defined as: \[ c_{ij} = (-1)^{i+j}\operatorname{det}(M_{ij}) \] where M_ij is a submatrix of A after removing the i-th row and the j-th column. Recall also that the determinant of a matrix A is a scalar value representing the sum of products of the elements in any row or column of A with their corresponding cofactors.
02

The adjugate formula

There is a useful relationship between a matrix A and its adjugate known as the adjugate formula: \[ A\cdot(\text{adj}(A)) = (\text{adj}(A))\cdot A = (\det(A))I \] where I is the identity matrix of the same size as A.
03

Calculate the determinant of adj(A)

Let's denote B = adj(A), so we have \[ A\cdot B = B\cdot A = (\det(A))I \] Now, we calculate the determinant of the left-hand side and the right-hand side of the first equation, respectively. Using the properties of determinants, we have: \[ \det(A\cdot B) = \det(A)\cdot\det(B) \hspace{1cm} (1) \\ \det((\det(A))I) = (\det(A))^n \hspace{1cm} (2) \] Since the left-hand sides of these two equations are directly related by the matrix multiplication, we can equate the right-hand sides: \[ \det(A) \cdot \det(B) = (\det(A))^n \hspace{3cm} (3) \]
04

Isolate the determinant of adj(A)

We want to find the determinant of the adjugate matrix B. To isolate \(\det(B)\), we just need to divide both sides of equation (3) by \(\det(A)\): \[ \det(B) = \frac{(\det(A))^n}{\det(A)} \] Simplifying the expression gives: \[ \det(B) = (\det(A))^{n-1} \]
05

Conclude

Now that we have found that \(\det(B) = (\det(A))^{n-1}\), we can substitute B = adj(A) back into the expression: \[ \det(\text{adj}(A)) = (\det(A))^{n-1} \] Thus, we have shown that the determinant of the adjugate of a nonsingular n x n matrix A is equal to the determinant of A raised to the power of n-1.

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Most popular questions from this chapter

Let \(A\) be a nonsingular \(n \times n\) matrix with a nonzero cofactor \(A_{n n},\) and set \\[ c=\frac{\operatorname{det}(A)}{A_{n n}} \\] Show that if we subtract \(c\) from \(a_{n n}\), then the resulting matrix will be singular.

A matrix \(A\) is said to be skew symmetric if \(A^{T}=-A .\) For example, \\[ A=\left(\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right) \\] is skew symmetric, since \\[ A^{T}=\left(\begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array}\right)=-A \\] If \(A\) is an \(n \times n\) skew-symmetric matrix and \(n\) is odd, show that \(A\) must be singular.

Use Cramer's rule to solve each of the following systems: (a) \(\quad x_{1}+2 x_{2}=3\) (b) \(2 x_{1}+3 x_{2}=2\) \(3 x_{1}-x_{2}=1\) \(3 x_{1}+2 x_{2}=5\) (c) \(2 x_{1}+x_{2}-3 x_{3}=0\) \(4 x_{1}+5 x_{2}+x_{3}=8\) \(-2 x_{1}-x_{2}+4 x_{3}=2\) (d) \(\quad x_{1}+3 x_{2}+x_{3}=1\) \(2 x_{1}+x_{2}+x_{3}=5\) \(-2 x_{1}+2 x_{2}-x_{3}=-8\) (e) \(x_{1}+x_{2}\) \(=0\) \(x_{2}+x_{3}-2 x_{4}=1\) \(x_{1}+2 x_{3}+x_{4}=0\) \(x_{1}+x_{2}+x_{4}=0\)

Show that if \(\operatorname{det}(A)=1,\) then \\[ \operatorname{adj}(\operatorname{adj} A)=A \\]

Let \(\mathbf{x}\) and \(\mathbf{y}\) be vectors in \(\mathbb{R}^{3}\) and define the skewsymmetric matrix \(A_{x}\) by \\[ A_{x}=\left(\begin{array}{rrr} 0 & -x_{3} & x_{2} \\ x_{3} & 0 & -x_{1} \\ -x_{2} & x_{1} & 0 \end{array}\right) \\] (a) Show that \(\mathbf{x} \times \mathbf{y}=A_{x} \mathbf{y}\) (b) Show that \(\mathbf{y} \times \mathbf{x}=A_{x}^{T} \mathbf{y}\)
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